Second Bertini theorem

 In this post, I try to prove Bertini theorem partly, since there is no public reference about the case that the linear system is base point free. The proof below is just an analogue of the one that was provided by Hartshorne long ago in his book "Algebraic geometry".

Theorem(second Bertini's theorem).  Let $X$ an algebraic variety over a closed field $k$ of characteristics $0$ and $dim X \geq 2$, let $\mathcal{L}$ be a linear system without fixed component on $V$, then for $D \in \mathcal{L}$ a general element of $\mathcal{L}$ we have $D_{sing} \subset X_{sing} \cup Bs \mathcal{L}$. Moreover, if $X$ is non-singular and a general element $D \in \mathcal{L}$ a big Cartier Divisor, then general elements of $|D|$ are non-singular and irreducible varieties.

  

Proof:

We will only prove the second statement. Since $D$ is a big Cartier divisor and $|D|$ a base point free linear system, then it could define a projective morphism $\phi_{D}: X \rightarrow \mathbb{P}(H^0(X,D))$ such that $\phi_D (X)$ is a projective sub-variety with the same dimension as $X$. By Stein factorization, we get that:

                                    $\phi_D: X \xrightarrow{\phi} X' \xrightarrow{p} \mathbb{P}(H^0(X,D))$


such that fibers of $\phi$ are connected and $p$ a finite morphism. Then for $H \in \mathbb{P}(H^0(V,D))$ and an integer $r$, we get: 


                                     $H^0(V,rD)=H^0(X,\phi^{*}(rp^{*}H) \cong H^0(V',\phi_{*}O_V \otimes O_{V'}(rp^*H) \cong H^0(V',rp^*H)$


Since $H$ is ample on $\mathbb{P}(H^0(X,D))$, then $p^*H$ is ample on $X'$, then we can take $r$ big enough such that $rp^*H = O_{X'}(rD) $ be very ample divisor. Since $R^if_{*}(O_X \otimes f^{*}O_{X'}(-rp^*H)) = R^if_{*}(O_X) \otimes O_{X'}(-rp^*H)$, hence $R^i\phi_{*}(O_{X}(-rD)) = 0$. This implies that $H^i(X,-rD)=H^i(X',\phi_{*}(-rD))=H^i(X',O_{X'}(-r))$.  Thus by $H^1(X,-rD)=H^1(X',O_X'(-r))=0 $ by Lemma of Enriques-Severi-Zariski (corollary III.7.8 in [Har77]). Now we get $H^1(X, -rD) = 0$. Take the exact sequence: 

                                      $0 \rightarrow O_X(-rD) \rightarrow O_X \rightarrow O_{rD} \rightarrow 0$.

 

Taking cohomology, we get that:


                                       $H^0(X,O_X) \rightarrow H^0(D, O_{rD}) \rightarrow 0$


Since $H^0(X,O_X) = k$, then $H^0(D, O_{rD}) = k$ and hence $D$ is connected. We know that $D$ is also non-singular, so if $D$ is not irreducible, then $D = D_1 + D_2 +\cdots+D_k$ where some $D_i$ and $D_j$ have non empty intersection, which means at at point it must have multiplicity 2 on $D$, contradicted to the smoothness property.


REFERENCE:

 [Har77] Robin Hartshorne. Algebraic Geometry (Graduated Text in Mathematics) 1st Edition 1977.

   

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